Hypergeometric card question I have a set of 60 things. Of those things, 24 belong to one type, 8 to another type, and 4 to a third type. If you select 7 things from the set, what's the probability of getting at least one of each together?
 A: One way to solve this is with inclusion-exclusion. It is easy to count the $7$-tuples which definitely miss some set of types, which might or might not miss any of the others. This means we can evaluate the terms of the following:
$\#$ ways to include all types
$$= \sum_{S \subset \lbrace 1,2,3 \rbrace} (-1)^{|S|} (\# 7-\text{tuples missing types in}~ S)$$
$ = {60 \choose 7} $
$-{60-24 \choose 7}-{60 - 8 \choose 7} -{60-4 \choose 7}$
$+{60-24-8 \choose 7}+{60-24-4 \choose 7}+{60-8-4 \choose 7}$
$-{60-24-8-4 \choose 7}$
$ = 386,206,920 - 8,347680-...+73,629,072-346,104$
$ = 89,990,144$
Mathematica code:
b[vec_] := Binomial[60 - Total[vec], 7] * (-1)^Length[vec]
Total[Map[b,Subsets[{24,8,4}]]

89990144

So about $90$ million out of $386$ million ($23.3\%$) of the combinations contain at least one of each type. If all $7$-tuples are equally likely, then the probability of getting one of each is $23.3\%$.
There are other methods. You can determine the number of ways to split $7$ types among the $4$ so that the first $3$ types have a multiplicity of at least $1$. For example, there could be $3$ of type $1$, $1$ of type $2$, $1$ of type $3$, and $2$ others, and the number of $7$-tuples with this multiplicity of types is ${24 \choose 3}{8 \choose 1}{4 \choose 1}{60-24-8-4 \choose 2}$. The number of terms you get here is ${7 \choose 3} = 35$ instead of $8$, and each term would be a product of $4$ binomial coefficients, so I think it would be more complicated, but it should give the same answer. 
